Let $\mathcal{H}^s$ be the s-dimensional Hausdorff measure on $\mathbb{R}^n$. We showed that for all $s > 0$ this is a Borel regular measure on $\mathbb{R}^n$. Now in our lecture notes, we state that it is no Radon measure for $s < n$. As a Radon measure is a Borel regular measure which has finite measure for all compact subsets $K \subset \mathbb{R}^n$, this means that for all $s < n$ we can find such a compact subset with infinite measure.
I know the term of the Hausdorff dimension. Let $h := \dim_\mathcal{H}(K)<n$. It is clear to me, that for all $s < h$, the s-dimensional Hausdorff measure will not be Radon. However, why is this also the case for $h \leq s < n$? Do we take a different compact subset or is there a reason why $K$ has infinite measure for $s \geq h$?
Just take $K$ to be the closed unit ball in $\mathbb R^n$. It has Hausdorff dimension $n$ so for any $s < n$ we have $\mathcal H^s(K) = \infty$ showing that $\mathcal H^s$ is not a Radon measure on $\mathbb R^n$.
See for example the passage from Exercise 13 to 14 in these notes by Terence Tao.